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I am to check for max & min values for the given function $$f(x, y, z) = xy^2z^3$$ which is defined on $$M = \left\{x, y, z > 0, x+2y+3z=6\right\}$$

So.. what I did is:

$F(x, y, z, \gamma) = xy^2z^3 - \gamma(x+2y+3z-6)$ therefore $\begin{cases} \frac{\partial F}{\partial x} = y^2z^3 - \gamma = 0\\ \frac{\partial F}{\partial y} = 2xyz^3-2\gamma = 0\\ \frac{\partial F}{\partial z} = 3xy^2z^2 -3\gamma = 0\\ \frac{\partial F}{\partial \gamma} = -x-2y-3z+6 = 0\\ \end{cases}$

After the calculation I get the possible point is $P = (1, 1, 1)$. So I count the 2nd derivatives : $$\left[\begin{array}{ccc} \frac{\partial^2 f}{\partial x^2} = 0 &\frac{\partial^2 f}{\partial x\partial y} = 2yz^3 &\frac{\partial^2 f}{\partial x\partial z} = 3y^2z^2 \\ \frac{\partial^2 f}{\partial y \partial x} = 2yz^3 &\frac{\partial^2 f}{\partial y^2} = 2xz^3 &\frac{\partial^2 f}{\partial y\partial z} = 6xyz^2 \\ \frac{\partial^2 f}{\partial z \partial x} = 3y^2z^2 &\frac{\partial^2 f}{\partial z \partial y} = 6xyz^2 &\frac{\partial^2 f}{\partial z^2} = 6xy^2z \end{array}\right]$$

So with our point P it looks like : $$\left[\begin{array}{ccc} 0&2&3\\ 2&2&6\\ 3&6&6 \end{array}\right]$$

And here's where I got lost. $\det_{1} = 0$ therefore we have no idea whether there is an extremum or not, but Wolfram says there's a local maximum in $(1, 1, 1)$. How should I obtain it?

Amzoti
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darenn
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  • Skip this comment -- you wrote $F$ in your Hessian, but did use $f$... – colormegone Jun 05 '13 at 20:13
  • isnt it what i did? – darenn Jun 05 '13 at 20:15
  • Yes, it just isn't what you typed... but check your determinant again -- it isn't zero. – colormegone Jun 05 '13 at 20:16
  • How come its not? $det_{1} = \frac{\partial ^2 f}{\partial x^2} = 0$ – darenn Jun 05 '13 at 20:18
  • That's what I got: the second cofactor isn't 0 . – colormegone Jun 05 '13 at 20:19
  • $det_{1} = 0, det_{2} = -4, det_{3} = 12$ isnt it like that? – darenn Jun 05 '13 at 20:20
  • I'm getting the same matrix you do, but I'm not finding an example of a problem with similar results. The matrix has one positive and two complex eigenvalues (with negative real parts). The matrix doesn't appear to be "negative-definite", so this doesn't seem like the critical point ought to be a local maximum. – colormegone Jun 05 '13 at 20:59
  • I was reviewing the work and found a typo in the Hessian. However, it only changes $ \ \det_3 \ $ to 30 and the eigenvalues are now one real positive and two real negative, so there is no material change to be made in the remarks on this critical point. – colormegone Jun 06 '13 at 05:18

3 Answers3

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Upon poking around a bit more, it looks like having H = 0 (what you call $ \ \det_1 \ $ ) is one of those situations where the standard "second derivative test" for three variables can't be applied. We can fall back on substituting the constraint into the function to produce

$$h(y,z) \ = \ (6 - 2y - 3z) \cdot y^2 \cdot z^3 \ = \ 6y^2z^3 \ - \ 2y^3z^3 \ - \ 3y^2z^4 , $$

for which the discriminant is

[EDIT]

$$D \ |_{(1,1)} \ = \ [ \ h_{yy} \ h_{zz} \ - \ (h_{yz})^2 \ ] \ |_{(1,1)} $$

$$= \ [ \ (12z^3 - 12yz^3 - 6z^4)\cdot (36y^2z \ - 12y^3z \ - 36y^2z^2) $$ $$ - \ (36yz^2 - 18y^2z^2 - 24yz^3)^2 \ ] \ |_{(1,1)} $$

$$ = \ (-6)(-12) \ - \ (-6)^2 \ = \ 36 \ > \ 0 \ . $$

Since $ \ h_{yy} \ |_{(1,1)} = -6 \ < 0 \ , $ this identifies this critical point as a local maximum in $ \ x \ $ under the constraint.

The constraint surface is a bit peculiar, as this maximum sits atop a rather narrow "ridge" near a very deep "drop-off" (if one may discuss the "terrain" of the function). This may explain the difficulty in applying the usual test. Below are two views of $ \ x = h(y,z) \ $ .

enter image description here

[I am making this revision as I found an error in a coefficient of one of the second derivatives since my original posting. There was also a typo in the Hessian matrix in OP's post, but this proved to be immaterial to the overall problem.]

colormegone
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  • As I found out, when there are limitations put to the domain of a function, we dont have to check each of the determinant, just the determinant of the whole matrix, which is $>0$ therefore we've got an extremum here. Thanks for your work on the subject! – darenn Jun 18 '13 at 08:30
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You can use Lagrange multiplier method and here is the answer provided by CAS

$$ [x=6-3\,z,y=0,z=z,\lambda_{{1}}=0,x{y}^{2}{z}^{3}=0]$$ $$[x=-2\,y+6,y=y,z=0,\lambda_{{1}}=0,x{y}^{2}{z}^{3}=0]$$ $$[x=1,y=1,z=1,\lambda_{{1}}=1,x{y}^{2}{z}^{3}=1].$$

Mhenni Benghorbal
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  • The issue isn't getting the critical point (we managed that); it is characterizing the point (1,1,1) using the Hessian. – colormegone Jun 05 '13 at 20:56
  • @RecklessReckoner: I had not seen any one suggested this method. By the way, the title asking for max and min and this is a technique for finding them. – Mhenni Benghorbal Jun 05 '13 at 21:00
  • I'm saying that OP has already found the critical point; they state the point coordinates midway down the paragraph. (In fact, the method used is a variant of the Lagrange multiplier method: that is how $ \ F(x) \ $ has been defined using $ \ f(x) \ $ . What they are asking about is that when the Hessian matrix is evaluated there, it doesn't fit the conditions for the "second derivative test" in three variables. – colormegone Jun 05 '13 at 21:14
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Rewrite your constraint as $x+y+y+z+z+z=6$ and apply the am/gm inequality, yielding $xy^2z^3\le 1$ ,with equality only when $x=y=z=1$ . The critical point is thus an absolute maximum over the range.

Bill Kleinhans
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